Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 07-Elec-A5 Electronics — May 2016.
Closed-book; non-communicating calculator permitted; three hours. Five questions,
20 marks each — all five constitute a complete paper, so every question is
solved below. Op-amps are ideal with supply rails of ±15 V unless a
question states otherwise.
Reference texts (22-Elec-A5 Electronics): A. S. Sedra
& K. C. Smith, Microelectronic Circuits, 7th/8th ed. (Oxford)
— diodes (Ch. 4), MOSFETs and biasing (Ch. 5/7), op-amp circuits
(Ch. 2), signal generators / wave-shaping (Ch. 17), MOS logic inverters
(Ch. 14); R. C. Jaeger & T. N. Blalock,
Microelectronic Circuit Design (McGraw-Hill).
Check / figure note: every schematic below was redrawn from the printed figures of the paper. Where the
paper omits a numeric supply (Q1), the transfer curve is drawn to scale for an
illustrative VDD = 5 V and the symbolic result is
stated alongside.
Given. Op-amp A1 drives diode D1
into node X (= C ∥ R to ground), with
A1’s feedback taken after the diode (a “super-diode”);
A2 is a unity buffer delivering $v_{OUT}=v_X$ to
RL. $R=R_L=1\text{ k}\Omega$, $C=10\ \mu\text{F}$. Input burst
period $\approx2.5$ ms (four cycles per 10 ms).
Find. (a) $v_{OUT}(t)$; (b) the circuit’s function.
Figure 2a. Precision half-wave rectifier
(A1 + D1) charging C ∥ R,
buffered by A2.
Approach. Identify the super-diode as a precision (drop-free)
positive rectifier, find the hold/discharge time constant $\tau=RC$, compare it with the
input period, and draw the peak-following envelope.
Precision rectification, no diode drop. When $v_{IN}\gt v_X$,
A1 drives its output high, D1 conducts, and the
feedback loop forces $v_X=v_{IN}$ (the 0.7 V drop is divided out by the open-loop
gain). When $v_{IN}\lt v_X$, D1 blocks and A1
disconnects from node X.
Hold / discharge time constant. With the diode blocked, C
discharges through R:
$$\tau=RC=(1\text{ k}\Omega)(10\ \mu\text{F})=\boxed{10\ \text{ms}}$$
Compare with the signal period. Each burst holds four cycles in
10 ms, so $T\approx2.5$ ms. Because $\tau=4T$, between successive positive
peaks the output falls only by the factor $e^{-T/\tau}=e^{-0.25}=0.779$, i.e. a ripple
$$V_r\approx V_{pk}\frac{T}{RC}=0.25\,V_{pk}.$$
Envelope during the bursts. The output rides the positive peaks:
$\approx1$ V (rippling down to $\approx0.78$ V) during the first burst, then
$\approx2$ V (down to $\approx1.56$ V) during the second — a small
saw-tooth ripple on a flat top.
After the input stops. With no further peaks, C discharges
monotonically from $\approx2$ V toward 0 with $\tau=10$ ms (essentially gone in
$\sim5\tau=50$ ms). The output is one-sided (never negative).
(b) Function. The circuit is a precision positive
peak / envelope detector — it rectifies and tracks the amplitude
(envelope) of the AC input, the front end of an AM envelope demodulator or a peak-reading
AC voltmeter.
Figure 2b. Input tone-burst (grey) and
the peak-detected output (blue): flat tops at 1 V then 2 V with ≈25%
ripple, then an exponential decay (τ = 10 ms).
